栈数据结构和代码

用数组创建栈

//使用数组来创建栈
//这是一个简单的扑克牌游戏，计算机使用随机数和栈洗牌
//在洗好后，将牌发给四家，每家5张牌
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define MAXSTACK 100    //最大栈容量

//栈数据的存入
int stack[MAXSTACK];    //栈的数组声明
int top = -1;           //栈的顶端

int push(int value)
{
    if(top >= MAXSTACK)             //是否超过容量
    {
        printf("栈内容全满\n");     //存入失败
        return -1;
    }
    top ++;                         //栈指针加1
    stack[top] = value;             //存入栈
    
}

//栈数据的取出
int pop()
{
    int temp;
    if(top < 0)                     //是否栈是空
    {
        printf("栈内容是空的\n");
        return -1;                  //取出失败
    }
    temp = stack[top];              //取出数据
    top--;                          //栈指针减1
    return temp;                    //栈取出
}



//主程序：洗牌后，将牌发给四个人
//红心：数组 0 ~ 12
//方块：数组 13 ~ 25
//梅花：数组 26 ~ 38
//黑桃：数组 39 ~ 51
int main()
{
    int card[52];                   //扑克牌数组
    int pos;                        //牌代码
    int i,temp;
    long temptime;
    srand(time(&temptime) % 60);    //使用时间初始随机数
    for(i = 0;i < 52; i++)
        card[i] = 0;                //清除扑克牌数组
    i = 0;
    while(i != 52)                  //洗牌循环
    {
        pos = rand() % 52;          //随机数取值 0 ~ 51
        if(card[pos] == 0)          //是否未洗牌
        {
            push(pos);              //存此张牌进栈
            card[pos] = 1;          //设置此张牌洗过
            i++;                    //下一张牌
        }
    }
    printf("    1     2      3       4\n");
    printf("=============================\n");
    for(i = 0;i < 5;i++)            //发牌给四人的循环
    {
        temp = pop();                  // 取出栈数据
        printf(" [%c%2d] ",temp /13 + 3,temp % 13 +1);
        temp = pop();
        printf(" [%c%2d] ",temp /13 + 3,temp % 13 +1);
        temp = pop();
        printf(" [%c%2d] ",temp /13 + 3,temp % 13 +1);
        temp = pop();
        printf(" [%c%2d] ",temp /13 + 3,temp % 13 +1);
        printf("\n");
        
    }
    return 0;
}

 

用链表创建栈

#include <time.h>
#include <stdio.h>
#include <stdlib.h>

struct stack_node
{
    int data;
    struct stack_node *next;
};

typedef struct stack_node stack_list;
typedef stack_list* link;

link stack = NULL;

int push(int value)
{
    link new_node;
    
    new_node = (link)malloc(sizeof(stack_list));
    if(!new_node)
    {
        printf("内存分配失败！\n");
        return -1;
    }
    new_node->data = value;
    new_node->next = stack;
    stack = new_node;
}

int pop()
{
    link top;
    int temp;
    
    if(stack != NULL)
    {
        top = stack;
        stack = stack->next;
        temp = top->data;
        free(top);
        return temp;
    }
    else
        return -1;
}


int empt()
{
    if(stack == NULL)
    {
        return 1;
    }
    else
        return 0;
}

int main()
{
    int card[52];
    int pos;
    int i,temp;
    long temptime;
    srand(time(&temptime)%60);
    for(i = 0;i < 52;i++)
        card[i] = 0;
    i = 0;
    while(i != 52)
    {
        pos = rand()%52;
        if(card[pos]  == 0)
        {
            push(pos);
            card[pos] = 1;
            i++;
        }
    }
    printf("    1     2      3       4\n");
    printf("=============================\n");
    for(i = 0;i < 5;i++)            //发牌给四人的循环
    {
        temp = pop();                  // 取出栈数据
        printf(" [%c%2d] ",temp /13 + 3,temp % 13 +1);
        temp = pop();
        printf(" [%c%2d] ",temp /13 + 3,temp % 13 +1);
        temp = pop();
        printf(" [%c%2d] ",temp /13 + 3,temp % 13 +1);
        temp = pop();
        printf(" [%c%2d] ",temp /13 + 3,temp % 13 +1);
        printf("\n");

    }
    return 0;
}

 

修改程序将栈指针当做参数输入栈操作函数push() 和pop()，可以同时处理两个以上的栈

//这个程序将两个数组的内容分别存入个别的栈，
//完成后将两个栈元素全部取出并打印
#include <time.h>
#include <stdio.h>
#include <stdlib.h>

struct stack_node
{
    int data;
    struct stack_node *next;
};

typedef struct stack_node stack_list;
typedef stack_list* link;

link stack1 = NULL;
link stack2 = NULL;

link push(link stack,int value)
{
    link new_node;

    new_node = (link)malloc(sizeof(stack_list));
    if(!new_node)
    {
        printf("内存分配失败！\n");
        return NULL;
    }
    new_node->data = value;
    new_node->next = stack;
    stack = new_node;
}

link pop(link stack,int *value)
{
    link top;
  
    if(stack != NULL)
    {
        top = stack;
        stack = stack->next;
        *value = top->data;
        free(top);
        return stack;
    }
    else
        *value = -1;
}


int empty(link stack)
{
    if(stack == NULL)
    {
        return 1;
    }
    else
        return 0;
}

int main()
{
    int list1[6] = {1,2,3,4,5,6};
    int list2[6] = {7,6,9,4,3,0};
    int i,temp;
    
    for(i = 0;i < 6;i++)
    {
        stack1 = push(stack1,list1[i]);
        stack2 = push(stack2,list2[i]);
    }
    printf("原来的数组顺序(1): ");
    for(i = 0;i < 6;i++)
        printf("[%d]",list1[i]);
    printf("\n");
    printf("栈取出的顺序(1): ");
    while(!empty(stack1))
    {
        stack1 = pop(stack1,&temp);
        printf("[%d]",temp);
    }
    printf("\n");
    printf("原来的数组顺序(2): ");
    for(i = 0;i < 6;i++)
        printf("[%d]",list2[i]);
    printf("\n");
    printf("栈取出的顺序(2): ");
    while(!empty(stack2))
    {
        stack2 = pop(stack2,&temp);
        printf("[%d]",temp);
    }
    printf("\n");
    return 0;
}